Mississipi Counting Problem?

1. Find the number of ways the letters in Mississipi can be arranged in groups of three.
2. Find the general formula for a group g letters.
3. For any given word with l letters labeled from 1 to l and these l letters appear n1, n2, n3, ..., nl times respectively, find the total number of ways to arrange... show more 1. Find the number of ways the letters in Mississipi can be arranged in groups of three.

2. Find the general formula for a group g letters.

3. For any given word with l letters labeled from 1 to l and these l letters appear n1, n2, n3, ..., nl times respectively, find the total number of ways to arrange these letters in groups of g letters.

Don't do all of them if you don't want to.

Update: I think I'll have to rephrase #3. Hopefully you guys still get the idea.

Update 2: If this were the common combination problem, I wouldn't have asked...

Mississipi repeats its letters many times. What is the way to deal with this?

Update 3: Only just a group of 3.
MIS is one group of 3.

I want to know how many different ways we can do this using all the letters.

Update 4: - -, are you sure it works? I don't think it works for g = 11.
Also, what is your name? Calling people by dashes is inhumane.

Answers

Best Answer: EDIT
I carelessly typed in the wrong generating function (but have corrected it below). It shouldn't be just (1+x²+x³+x^4)²; I forgot the term (1+x)². Sorry about that!
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I love combinatorics.
For these questions, you need to use generating functions to solve it.

You have 1 M, 4 i's, 4 s's, and 1 p. You require the number of combinations of 3 letters from these letters.

The generating function in this case is
(1+x)² (1+x²+x³+x^4)²

The answer is the coefficient of the term x³ in the expansion of the above expression.

If you want a group of g letters where g is a positive integer, you require the coefficient of x^g in the expansion of
(1+x)² (1+x²+x³+x^4)²

This can be easily solved by using a program, but unfortunately I don't have one.

For the third one, you require the coefficient of x^g in the expansion of

Edit:
I tried to include a link to generating functions in wikipedia but I encountered error 999 when I tried to submit it... is there a solution?

Edit:
The first answer's combination formula works only when each letter is distinct. In that case you require the coefficient x^q in the expansion of (1+x)^n where n is the number of distinct letters and q is the number of letters you want in each combination. In that case it is of course nCq = n!/(q!(n-q)!), which is the binomial coefficient.

Edit:
Does anyone know of any free online calculator for evaluating the expansion of an expression like (x^3 + x^8 + x^10) (x^2 + x^4 + x^5 + x^7), or at least finding the coefficient of a specific term in the expansion of the multinomials?
Many thanks if there's one online because I don't have a program to do all the business of simplying...

Edit:
Of course it won't work for g = 11 because there are only 10 letters in "Mississipi".

Are these questions from a specific problem set (with/without solutions), or did you made these up?

Edit:
With the correct expansion of the generating function (x+1)² (1+x²+x³+x^4)², you should get 1 for g = 10.

Do you have a program to do all the calculations? I hope you do.

Edit:
Did your teacher use a different approach other than generating functions to do the first question? But don't tell me that his approach is basic enumeration of all the combinations! :-)

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Edit:

OK. In that case it is PERMUTATION instead of COMBINATION.

Since it is permutation, you want to find the coefficient of x³/3! in the expansion of

The coefficient of x^r is C(n,r) and this is the number of ways of choosing r things from n different things. This arises because in the multiplication of n brackets, (1+x)(1+x)(1+x)(1+x)....to n terms, we can choose the r brackets from which to take the x and (n-r) brackets from which to choose 1 in C(n,r) ways. So the coefficient of x^r gives the number of ways of selecting r things from n things. Thus if we have 3 A's, 2 B's and 5 C's and require the number of combinations of 4 letters that can be chosen from these letters, the generating function is:

and we can see that coefficient of x^4 is 11, so there are 11 combinations of 4 letters that can be made from the 10 letters available.

Exponential Generating Functions
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We could write the expansion of (1+x)^n in a different form.
If P(n,r) is the number of permutations of r things that can be made from n different things then we have:

(1+x)^n = P(n,0) + P(n,1).x + P(n,2)x²/2! + P(n,3).x³/3! + ...

and the coefficient of x^r/r! will be P(n,r)

This arises because the number of PERMUTATIONS of the x's (considered to be different from each other) that could be made from choosing r brackets to contribute an x and n-r brackets to contribute a 1, is indeed P(n,r). So in this type of generating function the coefficient of x^r/r! is the number of permutations that could be made by selecting r things from n things.

Thus if we have 3 A's, 2 B's and 5 C's and require the number of permutations of 4 letters that can be chosen from these letters, the generating function is:

(1+x+x²/2!+x³/3!) (1+x+x²/2!) (1+x+x²/2!+x³/3!+x^4/4!+x^5/5!)

Since we need only coeffients up to x^4 we could write this

(1+x+x²/2!+x³/3!) (1+x+x²/2!) e^x

= (1 + 2x + 2x² + 7x³/6 + 5x^4/12 + x^5/12) e^x

= 1 + 3x + 9x²/2 + 13x³/3 + 71x^4/24 + 181x^5/120 + ...

And we require the coefficient of x^4/4! = coefficient of x^4/24.

From above working we see that this is 71. So there will be 71 permutations of 4 letters taken from the 10 letters available.

Note that you can use e^x if the number of letters that you require is not greater than the number of letters available. In this problem we could use up to 5 letters C and only required 4 letters for each permutation.

The whole problem is much easier if for instance you had the following question. How many different permutations of 4 letters can be made from 5 A's, 6 B's and 7 C's?

The answer will be given by the coefficient of x^4/4! in the expansion of:

(1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...)³

= [e^x]³ = e^(3x)

= 1 + (3x) + (3x)²/2! + (3x)³/3! + (3x)^4/4! + (3x)^5/5! + ...

and the coefficient of x^4/4! is clearly 3^4 = 81

So there will be 81 permutations of 4 letters that can be made from the 18 letters available.

We could calculate this directly, since we have 3 letters to choose from for each of the 4 positions, giving 3^4 = 81 sequences.

There is a common problem stating questions of this kind. In this case, does "MISSISSIPPI" contain 4 letters or 11 letters? In combinatorial language, we would say it contains 4 varieties, and 11 items of those 4 varieties.

When you ask how many ways the letters can be arranged in groups of three, do you mean arrange the 11 items in groups of three, or arrange the 4 varieties in groups of three?

Either way, the answer is ZERO. You can't arrange 11 items in groups of 3 because 3 is not a factor of 11. Nor can you arrange 4 varieties in groups of 3 because 3 is not a factor of 4.

Maybe you meant to arrange the items in 3 groups of 3 and 1 leftover group of 2.

If so, then there still is a problem of whether order makes a difference in groups, and whether order of groups makes a difference. For example:
Should we count MIS-SIS-SIP-PI as different from SIS-MIS-SIP-PI ?
Should we count MIS-SIS-SIP-PI as different from SIM-SIS-SIP-PI ?

Questions of this kind are notoriously difficult to state in a precise unambiguous way.

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